Question

If r is the radius of first orbit, the radius of nth orbit of the H atom will be:
(A) $r{n}^{2}$
(B) $rn$
(C) $r/n$
(D) ${r}^{2}{n}^{3}$

Answer 1

Hint: The radius of any orbit of any element in the periodic series is calculated in reference with the radius of the first orbit of the hydrogen atom as it has only one orbit and so it is easier to calculate the radius of the orbit in reference to hydrogen.
Complete step by step answer:
The radius of the orbit of a nucleus or the Bohr radius is a physical constant that is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is given by:
${ a }_{ 0 } = \cfrac { { h }^{ 2 }{ \varepsilon }_{ 0 } }{ \pi m{ e }^{ 2 } } = 5.29 \times { 10 }^{ -11 } m$
where ${a}_{0}$ = Bohr radius
h = Planck's constant
${\varepsilon}_{0} $ = $8.854 x {10}^{-12} F/m$
m=mass of an electron
and e is the electronic charge
If the radius r of the first orbital any element is given then the radius of the ${n}^{th}$ orbital of that element is given by:
${ r }_{ n }\ = \cfrac { r { n }^{ 2 } }{ Z }$ ---(1)
where ${ r }_{ n }$ = radius of ${n}^{th}$ orbital
n = the orbital for which the radius has to found out
Z = atomic number of the element
Now, let us find out the radius of the ${n}^{th}$ orbital of H atom. The atomic number of H atoms is 1. Therefore, Z=1. Now substituting the values in equation (1), we get
${ r }_{ n } = \cfrac { r { n }^{ 2 } }{ 1 }$
$\implies { r }_{ n } = r {n}^{2}$
Therefore, when r is the radius of the first orbit, then the radius of nth orbit of the H atom is $r {n}^{2}$. Hence option (a) is the correct option.

Note: In equation (1), if we substitute the general value of the radius of the first orbital of H atom i.e., $5.29 \times {10}^{-11} m$ and Z as 1 then the radius of the ${n}^{th}$ orbital that we will be getting would be $5.29 times {10}^{-11} x {n}^{2}$ m.